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Grade 12th passMechanics

Points at which the first derivative of a function equal zero may not all be minima; some may be maxima. Can you show, using the second derivative, that we have indeed found a minimum ofI?

Profile image of Rizwan Mahsood
9 Years agoGrade 12th pass
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Profile image of Askiitians Tutor Team
ApprovedApproved Tutor Answer1 Year ago

When analyzing the behavior of a function, particularly in calculus, we often look at the first derivative to find critical points where the function's slope is zero. However, identifying whether these points are minima, maxima, or neither requires further investigation, typically through the second derivative test. Let’s break this down step-by-step to clarify how the second derivative can help us confirm a minimum.

Understanding Critical Points

Critical points occur where the first derivative of a function, denoted as f'(x), equals zero or is undefined. At these points, the function may change from increasing to decreasing or vice versa. However, not all critical points are minima; some could be maxima or even points of inflection.

The Role of the Second Derivative

The second derivative, f''(x), provides insight into the concavity of the function. By evaluating the second derivative at a critical point, we can determine the nature of that point:

  • If f''(x) > 0: The function is concave up at that point, indicating a local minimum.
  • If f''(x) < 0: The function is concave down, suggesting a local maximum.
  • If f''(x) = 0: The test is inconclusive, and further analysis is needed.

Example to Illustrate the Concept

Let’s consider a specific function, f(x) = x^3 - 3x^2 + 4. We want to find the critical points and determine if they are minima or maxima.

Step 1: Find the First Derivative

First, we calculate the first derivative:

f'(x) = 3x^2 - 6.

Setting this equal to zero gives:

3x^2 - 6 = 0
x^2 = 2
x = ±√2.

Step 2: Evaluate the Second Derivative

Next, we find the second derivative:

f''(x) = 6x.

Now, we evaluate f''(x) at our critical points:

  • For x = √2:
    f''(√2) = 6√2 > 0, indicating a local minimum.
  • For x = -√2:
    f''(-√2) = -6√2 < 0, indicating a local maximum.

Conclusion of the Analysis

Through this process, we see that the critical points we found using the first derivative are indeed classified correctly using the second derivative. The point at x = √2 is a local minimum, while x = -√2 is a local maximum. This method of using the second derivative is a powerful tool in calculus for confirming the nature of critical points and ensuring that we accurately identify minima and maxima in functions.